Max. distance a clock can travel

Let us assume i have a clocks of 10KHz & 10MHz...
I am transmitting them through two individual cables, which signal can travel long distance and why?

Depends of cable characterists, like : wire width, wire geometry, pair distance, insulating matherial used, etc...
The main effects are :

Skin effect
Dielectric effect

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I have a similar cable for transmitting both & is of same length

The higher frequencies will be attenuated more because of capacitance between the conductors (near end cross talk) and greater skin effect losses.

10Khz under any situation will travel more than 10MHz, due to attenuation property

I understand, that aren't interested in the details, how to transmit 10 MHz over maximum distance.

But in any case, 10 kHz can be transmitted over longer distance, assuming identical conditions. Besides attenuation, dispersion will cause waveform distortion. Transmission line effects, particularly signal reflections, will restrict the distance, if no termination is provided.

Skin depth reduces as frequency increases which is acting in favour of lower frequencies....
But how can we explain capacitance?

---------- Post added at 12:47 ---------- Previous post was at 12:45 ----------

@sivaram

I feel by attenuation property you mean skin depth.... or any other?

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this is what i mean.. this should clear all your doubts regarding attenuation of larger frequency over requency small frequency

1. skin depth
2. capacitance
3. dispersion
4. reflections

Effect of skin depth is clear on frequencies.
What is the effect of other 3 factors on frequency?

Capacitance is frequency independant, but it's influence is increasing at higher signal frequencies. This is the case, if the cable is not operated as a terminated transmission line. Then it acts as a load capacitance with increasing load currents towards longer cables and higher frequencies.

Dispersion implies signal edges getting smeared with increasing cable length. Square waves get rounded. At a certain point, the high frequency square wave can't be reproduced any more. Strictly spoken, the fundamental frequency will be still recoverable, so the original waveform can be reconstructed. But existing signal receivers are possibly not designed for.

For an unterminated cable, the strongest signal reflections can be expected to travel end-to-end. So at the receiver, we'll see a series of signal "echos" with a distance of double the cable flight time, decreasing due to cable attenuation and imperfect reflection. At low frequencies, the echo series will have vanished between signal edges and can be easily filtered. At high frequencies, echos are superimposing with next signal edges, and time or frequency domain filtering can't be applied.

As Mr.dharanidhar said
1. skin depth
2. capacitance
3. dispersion
4. reflections

all are factors in similor conditions(Same voltage & current) for 10KHz & 10MHz... the lowest freaquency signal(10KHZ) can travel longer distances

Thank you very much for the information...
Now getting back all the basics of communication..

Also cable radiation, most screened cables do not have a complete metallic tube as the outer sheath, so signal literally leaks out of the cable between the braiding. This is why the cable used for satellite down feeds (.5 -> 1Ghz) have two layers of copper foil. This loss of the output is mopped up in "cable attenuation".
if the cable is made properly, its charecteristic impedance will vary down its length casing reflections. Rhode & Scwarzt who make high precision rf measuring kit use a relatively large diameter cable with desi fix connectors, with a cable of 1" diam. it is more likely to be spot on then a .2" diam cable. Reflections will again cause a lower output level , mopped up as cable attenuation.
Both the above effects worse at 10Mhz then 10Khz.
Frank

How can we explain leakage effects in cables w.r.t frequency?

For a given size "hole" in the screening, the higher the frequency, the more power leaks out. If the hole is half wavelength long, it will be a very effective aerial, i.e. all the power will radiate out of the "slot/hole".
Frank